What is the integral of (xln(x^2+4))/(x^2+4) ?

The integral of (xln(x^2+4))/(x^2+4) is 1/4 ln^2(x^2+4)+C

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integral of (xln(x^2+4))/(x^2+4)

\int \frac{x ln ( x ^{2} + 4 )}{x ^{2} + 4} d x

\frac{1}{4} ln^{2} (x^{2} + 4) + C

Solution steps

\int \frac{x ln ( x ^{2} + 4 )}{x ^{2} + 4} d x

Apply u-substitution

: \int \frac{ln ( u + 4 )}{2 ( u + 4 )} d u

= \int \frac{ln ( u + 4 )}{2 ( u + 4 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{ln ( u + 4 )}{u + 4} d u

Apply u-substitution

: \int \frac{ln ( v )}{v} d v

= \frac{1}{2} \cdot \int \frac{ln ( v )}{v} d v

Apply u-substitution

: \int w d w

= \frac{1}{2} \cdot \int w d w

Apply the Power Rule

: \frac{w ^{2}}{2}

= \frac{1}{2} \cdot \frac{w ^{2}}{2}

Substitute back

= \frac{1}{2} \cdot \frac{ln ^{2} ( x ^{2} + 4 )}{2}

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( x ^{2} + 4 )}{2} : \frac{1}{4} ln^{2} (x^{2} + 4)

= \frac{1}{4} ln^{2} (x^{2} + 4)

Add a constant to the solution

= \frac{1}{4} ln^{2} (x^{2} + 4) + C

\int x^{2} x^{3} + 37 d x

\int \frac{x + 2}{x - 1} d x

f (x) = 7 x + 2, at x = 2

(0, 1), (5, 4)

(21 + x^{2}) y^{^{'}} - 2 x y = 0

What is the integral of (xln(x^2+4))/(x^2+4) ?
The integral of (xln(x^2+4))/(x^2+4) is 1/4 ln^2(x^2+4)+C